Question: $ 16^{-\frac{1}{2}}$
$= \left(\dfrac{1}{16}\right)^{\frac{1}{2}}$ Figure out what goes in the blank: $\Big(? \Big)^{2}=\dfrac{1}{16}$ Figure out what goes in the blank: $\Big({\dfrac{1}{4}}\Big)^{2}=\dfrac{1}{16}$ So $16^{-\frac{1}{2}}=\left(\dfrac{1}{16}\right)^{\frac{1}{2}}=\dfrac{1}{4}$